Calculate the maximum deceleration of a car that is heading down a 14.5° slope (
ID: 1438451 • Letter: C
Question
Calculate the maximum deceleration of a car that is heading down a 14.5° slope (one that makes an angle of 14.5° with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the static coefficient of friction is involved--that is, the tires are not allowed to slip during the deceleration.
(a) on dry concrete
m/s2
(b) on wet concrete
m/s2
(c) on ice, assuming that µs = 0.100, the same as for shoes on ice
m/s2
Explanation / Answer
In this case
Ff = mg*sin A - uN = m*a
N = mgcos A
total F = 0
ma = m*g*sinA - umg*cos A
deceleration a = (g*sin A - g*us*cos A)
A = 14.5 deg
us for rubber on dry concrete = 1.0
us for rubber on wet concrete = 0.7
us for ice = 0.1
A.
a = 9.81*sin 14.5 deg - 9.81*1.0*cos 14.5 deg = -7.041 m/sec^2
B.
a = 9.81*sin 14.5 deg - 9.81*0.7*cos 14.5 deg = -4.192 m/sec^2
C.
a = 9.81*sin 14.5 deg - 9.81*0.1*cos 14.5 deg = 1.506 m/sec^2
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